devin553344 8 Posted July 30, 2020 Author Report Share Posted July 30, 2020 I've had enough for this: wont speak for anyone else.Thanks for your time. Quote Link to post Share on other sites

devin553344 8 Posted August 2, 2020 Author Report Share Posted August 2, 2020 (edited) I've adjusted some of the equations in the new theory, and brought together all the best equations that all fit together like a puzzle to make sense of the different ideas.Â Gravitation is now described as a wave strong force at the wavelength similar to the particle strong force I've previously described. The following is the Compton wavelengths of the strong force energy divided by the particle energy:Â rF/rC= 4Ď€exp(2)Â Where rF is the strong force Compton wavelength at the wavelength, rC is the Compton wavelength of the particle.Â These wavelengths go into the wave strong force at the wavelength to describe the gravitational reduction. Since I've described the particles as have a logarithmic strain of the wavelength of the electron divided by proton and vice versa, this will be a modifier to the gravitational curvature:Â Gme^2/re = mec^2 * rp/re *1/(rF/rC * exp(rF/rC)Â Gmp^2/rp = mpc^2 * re/rp *1/(rF/rC * exp(rF/rC)Â Where G is the gravitational constant, me is the mass of the electron, re is the wavelength of the electron, c is the speed of light, rp is the wavelength of the proton, mp is the mass of the proton.Â And if this wave strong force exists there should be proof in Planck's constant. There is a charge plane equation and a sphere to sphere both at 1/2 the wavelength to the forth for pressure or energy per meter cubed (although I've shown energy times meters so simply divide across by radius to the forth to get pressure). This appears to support a wave (charge planes) and particle (point charges) duality:Â hc/exp^4(1/2) = 16e^2/(2Îµ) + 16e^2/(4Ď€Îµ)Â Where h is the Planck constant, c is the speed of light, e is the elementary charge and Îµ is the permittivity of free space. The factor of 2 per charge might be a dipole moment of the charge difference.Â The other electromagnetic strong force is 5 dimensional:Â hc/(8/3Ď€^2exp(4)) = 3/5 * Ke^2Â Where K is the electric constant.Â I've updated the PDF file in the OP at the beginning of this thread. Edited August 3, 2020 by devin553344 Quote Link to post Share on other sites

devin553344 8 Posted August 5, 2020 Author Report Share Posted August 5, 2020 (edited) I updated the PDF file in the OP to reflect the 3 & 5 dimensional radii that match mc^2, the 3 dimensional controls the magnetic moment of the proton while the 5 dimensional determines the charge radius. The magnetic moment is then:Â mc^2 = mc^2/(4Ď€ ru^2/(4*rp^2) * exp(2*ru/(2*rp))))Â ru = 5.951 894E-16 metersÂ u = 1/2 e ru c = 1.429 408E-26 A * m^2Â Where m is the mass of the proton, c is the speed of light, ru is the magnetic moment radius at the speed of light, rp is the proton wavelength, u is the magnetic moment of the proton, e is the elementary charge.Â The charge radius is then:Â mc^2 = mc^2/(8/3Ď€^2 rC^2/(16*rp^2) * exp(4*rC/(2*rp))))Â rC = 8.470 237E-16 metersÂ Where rC is the charge radius. See: https://en.wikipedia.org/wiki/Proton_radius_puzzle Edited August 5, 2020 by devin553344 Quote Link to post Share on other sites

devin553344 8 Posted August 7, 2020 Author Report Share Posted August 7, 2020 (edited) I've revised the wave-particle duality portion of the theory which modifies the strain energy equations. Now I derive the energy of the proton and electron from the electric energy using the charge radius which for the proton is 8.45E-16 meters.Â After reviewing the wave-particle duality ideas I've come to the conclusion that the following must be true:Â E/E0 = Î»/Î»0 = Ďµ^2/2Â E is the increased energy, E0 is Young's modulus pressure times volume or energy, Î» is the decreased wavelength, Î»0 is the original wavelength, Ďµ is the strain. The strain must be a strain of the wavelengths involved, and I have used only 5 dimensional strains for the description of electromagnetic deltas.Â I've used the electron/proton wavelength to derive two solutions to this new wave-particle idea. One for the electron (point particle) and one for the proton (particle with a radius). I updated the gravitation to match these strain values as a leak of the matter curvature.Â I've updated the PDF file in the OP but will probably not spell the equations out here. Edited August 7, 2020 by devin553344 Quote Link to post Share on other sites

devin553344 8 Posted August 8, 2020 Author Report Share Posted August 8, 2020 OK I've finally worked out the details. I've updated the PDF in the OP and will post the various equations shortly. Quote Link to post Share on other sites

devin553344 8 Posted August 8, 2020 Author Report Share Posted August 8, 2020 (edited) The electron is now defined as a point particle with a radius of the Planck length due to it's radius strain:Â mec^2 = 1/2 * 2/3 * Ke^2/(hc) * (ln(4Ď€LP/re))^2Â Where me is the mass of the electron, c is the speed of light, K is the electric constant, e is the elementary charge, h is Planck's constant, LP is the Planck length, re is the wavelength of the electron.Â Now to strain out the system and access the electron curvature thru gravitation I have:Â 8Ď€Gme^2/re = mec^2 * exp(2ln(LP/re))Â Where G is the gravitational constant.Â The proton gravitation is now (which is about the same value as the neutron):Â 8Ď€Gmp^2/rp = mpc^2 * exp(2ln(LP/re)Â + 2ln(re/rp))Â Where mp is the mass of the proton, rp is the proton wavelength.Â The proton is comprised of two super charged positrons and one electron:Â mpc^2 = 3/2 * mec^2 * (4ln(rp/re) + ln(a0))^2Â Where a0 is the fine structure constant.Â And now the neutron is comprised of two supercharged positrons and two electrons with a reversed strain direction:Â mnc^2 = 4/2 * mec^2 * (4ln(rn/re) - ln(3/5*a0/(2Ď€)))^2Â Where mn is the neutron mass, rn is the neutron wavelength.Â Cheers. Edited August 9, 2020 by devin553344 Quote Link to post Share on other sites

devin553344 8 Posted August 9, 2020 Author Report Share Posted August 9, 2020 (edited) Oops... small typo in the last post, but I corrected it, the electron derivation should have been a natural logarithm of the radius ratio. Also I found this relationship that might allow for pair production more easily, and this new strain is more accurate:Â ln(4Ď€LP/re) = 6ln(rp/re) + ln(4/9 * a0)Â Again where LP is the Planck length, re is the electron wavelength, rp is the proton wavelength, a0 is the fine structure.Â So that the electron definition might also be described as:Â mec^2 = 1/2 * 2/3 * Ke^2/(hc) * (6ln(rp/re) + ln(4/9 * a0))^2 Edited August 9, 2020 by devin553344 Quote Link to post Share on other sites

devin553344 8 Posted August 10, 2020 Author Report Share Posted August 10, 2020 (edited) Since I've been using the fine structure similar to a radius, then it should relate to radius ratios somehow. I looked and found this simple solution that relates the ratio half volume to the electric binding energy divided by Planck's constant:Â ( 1/2 * 4/3 * Ď€ * rp^3/re^3 )^1/3 = 3/5 * a0/(2Ď€)Â Where rp is the proton wavelength, re is the electron wavelength and a0 is the fine structure.Â And if there is a 3 dimensional version then there should be a 5 dimensional version since this is an electromagnetic equation. I find the following is more accurate:Â ( 1/2 * ((8/15 * Ď€^2)^2 * rp^5/re^5 * RZ(5))/(4/3 * Ď€) )^1/5 = 3/5 * a0/(2Ď€)Â Where RZ(5) is Riemann zeta 5 and where 8/15 * Ď€^2 * r^5 is a 5 dimensional volume n-sphere, see: (https://en.wikipedia.org/wiki/N-sphere) Edited August 10, 2020 by devin553344 Quote Link to post Share on other sites

devin553344 8 Posted August 14, 2020 Author Report Share Posted August 14, 2020 (edited) I've modified the theory as I believe that the strain for the gravitation should be squared similar to the energy strain. The following equations now fit. Please note the use of strain symbol and permittivity symbol. I use a factor of 16 to demonstrate the pair production examples:Â The electron:Â mc^2 = 16 * Îµ/2 * K^2e^2/r * 1/2 * Ďµ^2 = 8.187E-14 JoulesÂ 3/5 * G4m^2/r = mc^2 * exp(-2Ďµ) = 5.474E-59 JoulesÂ Where m is the mass of the electron, c is the speed of light, Îµ is the permittivity of free space, K is the electric constant, e is the elementary charge, r is the wavelength of the electron, Ďµ is the strain of the electron, G is the gravitational constant, exp is the natural exponent function.Â The proton:Â mc^2 = 2/9 * 16 * Îµ/2 * K^2e^2/r * 1/2 * Ďµ^2 = 1.503E-10 JoulesÂ Gm^2/r = mc^2 * exp(-2Ďµ) = 8.837E-49 JoulesÂ Where m is the mass of the proton, r is the reduced wavelength of the proton, Ďµ is the strain of the proton.Â The charge radius of the proton might be described via the ratio of a wave's linear nature deformed by two pi:Â r = 4c/(2Ď€v) = 8.412E-16 metersÂ And that unifies the strong force of mc^2 with gravitation and electromagnetic. Again the strong force I've defined as:Â mc^2/(4Ď€ * r^2/d^2 * exp(2 * r/d))Â Where m is the mass of the proton, r is the distance between nucleons, d is the wavelength of the proton.Â Please let me know if you find any typos, I've checked all the equations for accuracy in MS Excel.Â The strain of the electron calculates to 52.0094433669536 and the strain of the proton calculates to 44.0147792906283.Â And the strong force for deuteron calculates to -3.56478752359958E-13 Joules for two nucleons. Edited August 14, 2020 by devin553344 Quote Link to post Share on other sites

devin553344 8 Posted August 18, 2020 Author Report Share Posted August 18, 2020 I adjusted the strong force and made it an adjacent vacuum to the pressure of the logarithmic strain, it creates a charge like region outside the charge radius of the proton. I updated the PDF in the OP. Quote Link to post Share on other sites

devin553344 8 Posted August 21, 2020 Author Report Share Posted August 21, 2020 For anyone that's interested I updated the PDF in the OP with everything that makes sense. It's a curvature model of matter and charge. Thanks for your time :) Cheers. Quote Link to post Share on other sites

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